2 x 1 x 2 integral Light blue region feasible region of LP relaxation Black

2 x 1 x 2 integral light blue region feasible region

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Lizhi Wang ([email protected]) IE 534 Linear Programming October 31, 2012 6 / 21
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Lizhi Wang ([email protected]) IE 534 Linear Programming October 31, 2012 7 / 21
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max 0 s . t . a 1 x 1 + a 2 x 2 b x 1 , x 2 0 . Lizhi Wang ([email protected]) IE 534 Linear Programming October 31, 2012 8 / 21
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Totally unimodular matrix If the A matrix has a special property (totally unimodular) and b is integral, then ( A, b ) = ( ˜ A, ˜ b).A matrix istotally unimodular(TU) if the determinant of everysquare submatrix is0,1, or-1.IA subset of columns and rows of a matrix is a submatrix.IA11A14A31A34is a submatrix ofA11A12A13A14A21A22A23A24A31A32A33A34IfAis TU, thenAIis also TU. (There are many similarproperties).The node-arc incidence matrix of a directed graph is TU. Lizhi Wang ([email protected]) IE 534 Linear Programming October 31, 2012 9 / 21
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Example - 4 6 1 1 is not TU, because det - 4 6 1 1 = - 10 , det [ - 4] = - 4 , and det [6] = 6 . 1 - 1 0 0 0 0 0 0 - 1 0 1 - 1 0 1 0 0 0 1 - 1 1 1 0 1 0 0 0 0 0 - 1 - 1 0 1 0 0 0 0 0 0 - 1 - 1 is TU, because it’s a node-arc incidence matrix. Lizhi Wang ([email protected]) IE 534 Linear Programming October 31, 2012 10 / 21
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Node-arc incidence matrix Suppose a graph has n nodes and m arcs The node-arc incidence matrix is an n × m matrix For each arc k = ( i, j ) , A ik = 1 and A jk = - 1 Other elements are zeros - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 A B C D E F 1 1 0 0 0 0 0 - 1 - 1 0 1 0 0 0 0 1 0 - 1 - 1 - 1 1 1 0 0 0 0 0 1 0 0 - 1 0 0 0 0 0 0 -
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